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Current Question (ID: 20024)

Question:

$\text{Let a total charge } 2Q \text{ be distributed in a sphere of radius } R, \text{ with the charge density given by } \rho(r) = kr, \text{ where } r \text{ is the distance from the centre.}$ $\text{Two charges } A \text{ and } B, \text{ of } -Q \text{ each, are placed on diametrically opposite points, at equal distance, } a \text{ from the centre.}$ $\text{If } A \text{ and } B \text{ do not experience any force, then:}$

Options:

1. $a = 8^{-1/4}R$
2. $a = 2^{-1/4}R$
3. $a = \frac{3R}{2^{1/4}}$
4. $a = \frac{R}{\sqrt{3}}$

Solution:

$\text{Hint: Force due to shell equal to force due to other negative charge.}$ $\text{Total charge } = 2Q$ $\text{Charging density } \rho = kr$ $\text{Radius } = R$ $\int dq = \int_0^R kq \cdot 4\pi r^2 \, dr$ $2Q = k4\pi \int_0^R r^3 \, dr$ $2Q = \frac{k4\pi R^4}{4}$ $k = \frac{2Q}{\pi R^4} \quad \text{(1)}$ $F_{BA} = F_{\text{sphere}}$ $\text{Force on charge } B \text{ due to element}$ $dF = \frac{k(dq)Q}{a^2} = \frac{kq(K4\pi r^3)}{a^2} \, dr$ $F = \frac{kQK4\pi}{a^2} \int_0^a r^3 \, dr = \frac{kQK4\pi a^2}{4}$ $F_{BA} = F_{\text{sphere}}$ $\frac{kQ^2}{(2a)^2} = kQ4Ka^2 = \text{By replace value of } K$ $\frac{Q^2}{4a^2} = \frac{2Q^2}{\pi R^2} \pi a^2$ $a^4 = \frac{R^4}{8}$ $a = R \cdot 8^{-1/4}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}